24. Δ(델타)결선

24. Δ(델타)결선

\(V_{L}=|\mathbf{V}_{ab}|=|\mathbf{V}_{bc}|=|\mathbf{V}_{ca}|\) (선전압)

\(V_{p}=|\mathbf{V}_{an}|=|\mathbf{V}_{bn}|=|\mathbf{V}_{cn}|\) (상전압)

\(V_{L}=\sqrt{3}V_{p}\), \(\mathbf{V}_{ab}=\sqrt{3}V_{p}\angle30^{\circ}\)

\(\displaystyle\mathbf{I}_{AB}=\frac{\mathbf{V}_{ab}}{\mathbf{Z}_{p}}\), \(\displaystyle\mathbf{I}_{BC}=\frac{\mathbf{V}_{bc}}{\mathbf{Z}_{p}}\), \(\displaystyle\mathbf{I}_{CA}=\frac{\mathbf{V}}{\mathbf{Z}_{p}}\) (\(\mathbf{Z}_{p}\)는 유도성 임피던스)

3상 전류들의 크기: \(I_{p}=|\mathbf{I}_{AB}|=|\mathbf{I}_{BC}|=|\mathbf{I}_{CA}|\)

선전류의 크기: \(I_{L}=|\mathbf{I}_{aA}|=|\mathbf{I}_{bB}|=|\mathbf{I}_{cC}|\)

\(I_{L}=\sqrt{3}I_{p}\)

\(\Delta\)결선	상전압=선전압	선전류>상전류
Y결선	선전압>상전압	선전류=상전류

Y결선에서 역률각이 \(\theta\)이면 한 상이 흡수하는 전력은 \(\displaystyle P_{p}=V_{p}I_{p}\cos\theta=V_{p}I_{L}\cos\theta=\frac{V_{L}}{\sqrt{3}}I_{L}\cos\theta\)이고 전체 전력은 \(P=3P_{p}=\sqrt{3}V_{p}I_{p}\cos\theta\)이다.

\(\Delta \)결선에서 역률각이 \(\theta\)이면 한 상이 흡수하는 전력은 \(P_{p}=V_{p}I_{p}\cos\theta=V_{L}I_{p}\cos\theta=V_{L}\frac{I_{L}}{\sqrt{3}}\cos\theta\)이고 전체전력은 \(P=3P_{p}=\sqrt{3}V_{p}I_{p}\cos\theta\)이다.

부하	상전압	선전압	상전류	선전류
Y	\(\mathbf{V}_{AN}=V_{p}\angle0^{\circ}\) \(\mathbf{V}_{BN}=V_{p}\angle-120^{\circ}\) \(\mathbf{V}_{CN}=V_{p}\angle-240^{\circ}\)	\(\begin{align*}\mathbf{V}_{AB}&=\mathbf{V}_{ab}=(\sqrt{3}\angle30^{\circ})\mathbf{V}_{AN}\\&=\sqrt{3}V_{p}\angle30^{\circ}\end{align*}\) \(\begin{align*}\mathbf{V}_{BC}&=\mathbf{V}_{bc}=(\sqrt{3}\angle30^{\circ})\mathbf{V}_{BN}\\&=\sqrt{3}V_{p}\angle-90^{\circ}\end{align*}\) \(\begin{align*}\mathbf{V}_{CA}&=\mathbf{V}_{ca}=(\sqrt{3}\angle30^{\circ})\mathbf{V}_{CN}\\&=\sqrt{3}V_{p}\angle-210^{\circ}\end{align*}\)	\(\displaystyle\mathbf{I}_{aA}=\mathbf{I}_{AN}=\frac{\mathbf{V}_{AN}}{\mathbf{Z}_{p}}\) \(\mathbf{I}_{bB}=\mathbf{I}_{BN}=\frac{\mathbf{V}_{BN}}{\mathbf{Z}_{p}}\) \(\displaystyle\mathbf{I}_{cC}=\mathbf{I}_{CN}=\frac{\mathbf{V}_{CN}}{\mathbf{Z}_{p}}\)	\(\displaystyle\mathbf{I}_{aA}=\mathbf{I}_{AN}=\frac{\mathbf{V}_{AN}}{\mathbf{Z}_{p}}\) \(\displaystyle\mathbf{I}_{bB}=\mathbf{I}_{BN}=\frac{\mathbf{V}_{BN}}{\mathbf{Z}_{p}}\) \(\displaystyle\mathbf{I}_{cC}=\mathbf{I}_{CN}=\frac{\mathbf{V}_{CN}}{\mathbf{Z}_{p}}\)
\(\Delta\)	\(\mathbf{V}_{AB}=\mathbf{V}_{ab}=\sqrt{3}V_{p}\angle30^{\circ}\) \(\mathbf{V}_{BC}=\mathbf{V}_{bc}=\sqrt{3}V_{p}\angle-90^{\circ}\) \(\mathbf{V}_{CA}=\mathbf{V}_{ca}=\sqrt{3}V_{p}\angle-210^{\circ}\)	\(\mathbf{V}_{AB}=\mathbf{V}_{ab}=\sqrt{3}V_{p}\angle30^{\circ}\) \(\mathbf{V}_{BC}=\mathbf{V}_{bc}=\sqrt{3}V_{p}\angle-90^{\circ}\) \(\mathbf{V}_{CA}=\mathbf{V}_{ca}=\sqrt{3}V_{p}\angle-210^{\circ}\)	\(\displaystyle\mathbf{I}_{AB}=\frac{\mathbf{V}_{AB}}{\mathbf{Z}_{p}}\) \(\displaystyle\mathbf{I}_{BC}=\frac{\mathbf{V}_{BC}}{\mathbf{Z}_{p}}\) \(\displaystyle\mathbf{I}_{CA}=\frac{\mathbf{V}_{CA}}{\mathbf{Z}_{p}}\)	\(\displaystyle\mathbf{I}_{aA}=(\sqrt{3}\angle-30^{\circ})\frac{\mathbf{V}_{AB}}{\mathbf{Z}_{p}}\) \(\displaystyle\mathbf{I}_{bB}=(\sqrt{3}\angle-30^{\circ})\frac{\mathbf{V}_{BN}}{\mathbf{Z}_{p}}\) \(\displaystyle\mathbf{I}_{cC}=(\sqrt{3}\angle-30^{\circ})\frac{\mathbf{V}_{CN}}{\mathbf{Z}_{p}}\)

상당전력: Y부하: \(\sqrt{3}V_{L}I_{L}\cos\theta\), \(\Delta\)부하: \(\sqrt{3}V_{L}I_{L}\cos\theta\) (\(\cos\theta\)는 부하의 역률)

Y결선과 \(\Delta\)결선 사이의 평형 3상 부하의 관계: \(\displaystyle\mathbf{Z}_{Y}=\frac{\mathbf{Z}_{\Delta}}{3}\) 

참고자료:

Engineering Circuit Analysis 8th edition, Hayt, Kemmerly, Durbin, McGraw-Hill